For example, consider $f(n) = n^3\;$ vs $\;g(n) = (n-2)^3$.
These two functions are considered to be asymptotically equivalent $f(n)\sim g(n)$ since $\lim_{n\to \infty}\dfrac{f}{g}=1$.
But $|f-g|$ blows up at the rate of $O(n^2)$.
So the equivalence class defined by $f\sim g$ is a much bigger class than $|f-g|\to0$.
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Compare that to $f(n) = e^{n^3 + O(n)}\;$ vs $\;g(n) = e^{n^3}$.
Then $f$ clearly dominates $g$ multiplicatively, ie. $\lim_{n\to \infty}\dfrac{f}{g}=\infty$
So, multiplicative notion of equivalence is not very good for things like this.
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So I am wondering if there is a notion of different tiers of equivalence of two growths.
Say level 0 equivalence is defined as $\dfrac{f}{g}\to 1$,
Then level -1 equivalence is $ \dfrac{\exp f}{\exp g}\to 1$ which is same as $f-g\to 0$
and level 1 equivalence would be $\dfrac{\log{f}}{\log{g}}\to 1$, and under this level of equivalence, $f(n) = e^{n^3 + O(n)}$ and $g(n) = e^{n^3}$ would be considered the same level of growth.
and of course we can iteratively define this for all integers as $\dfrac{\log^k f}{\log^k{g}} \to 1$ for all positive integers and similarly flipping log and exp for negative integers.
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The main motivation behind this is, when dealing with asymptotic behaviors of very fast growth, we frequently write stuff like $O\left( n^{\log{n} + O(1)} \right)$ since $n^{O(1)}$ plays a nontrivial part with respect to the level 0 growth, but in the grand scheme of how fast the function is growing, the polynomial part plays very small part to the overall growth.
Then of course we can generalize the notion to work for Big-O, Big-Omega, etc.